1 Lecture 18 Closure Properties of Language class LFSA –Remember ideas used in solvable languages unit –Set complement –Set intersection, union, difference, symmetric difference

2 LFSA is closed under set complement If L is in LFSA, then L c is in LFSA Proof –Let L be an arbitrary language in LFSA –Let M be the FSA such that L(M) = L M exists by definition of L in LFSA –Construct FSA M’ from M –Argue L(M’) = L c –There exists an FSA M’ such that L(M’) = L c –L c is in LFSA

3 Visualization Let L be an arbitrary language in LFSA Let M be the FSA such that L(M) = L M exists by definition of L in LFSA Construct FSA M’ from M Argue L(M’) = L c L c is in LFSA LcLc L LFSA FSA’s M M’

4 Construct FSA M’ from M What did we do when we proved that REC, the set of solvable languages, is closed under set complement? Construct program P’ from program P –P’ is basically the same program as program P –The only difference comes at the end –If P is about to return no, P’ returns yes instead –If P is about to return yes, P’ returns no instead Can we translate this to the FSA setting?

5 Construct FSA M’ from M M = (Q, , q 0, A,  ) M’ = (Q’,  ’, q’, A’,  ’) –M’ should say yes when M says no –M’ should say no when M says yes –How? Q’ = Q  ’ =  q’ = q 0  ’ =  A’ = Q-A

6 Example a a a b b b FSA M a a a b b b FSA M’ Q’ = Q  ’ =  q’ = q 0  ’ =  A’ = Q-A

7 Construction is an algorithm Set Complement Construction –Algorithm Specification Input: FSA M Output: FSA M’ such that L(M’) = L(M) c –Comments This algorithm can be in any computational model. –It does not have to be (and typically is not) an FSA These set closure constructions are useful. –More on this later Construction Algorithm FSA M FSA M’

8 Construction is an algorithm Your algorithm must give a complete specification of M’ in terms of M –Example: Let input FSA M = (Q, , q 0, A,  ) Output FSA M’ = (Q’,  ’, q’, A’,  ’) where –Q’ = Q –  ’ =  –q’ = q 0 –  ’ =  –A’ = Q-A When describing such constructions, I will often only focus on the critical component, in this case A’ = Q-A, but you must specify all components of M’ in terms of M. Construction Algorithm FSA M FSA M’

9 LFSA closed under Set Intersection Operation (also set union, set difference, and symmetric difference)

10 LFSA closed under set intersection operation Let L 1 and L 2 be arbitrary languages in LFSA Let M 1 and M 2 be FSA’s s.t. L(M 1 ) = L 1, L(M 2 ) = L 2 –M 1 and M 2 exist by definition of L 1 and L 2 in LFSA Construct FSA M 3 from FSA’s M 1 and M 2 Argue L(M 3 ) = L 1 intersect L 2 There exists FSA M 3 s.t. L(M 3 ) = L 1 intersect L 2 L 1 intersect L 2 is in LFSA

11 Visualization Let L 1 and L 2 be arbitrary languages in LFSA Let M 1 and M 2 be FSA’s s.t. L(M 1 ) = L 1, L(M 2 ) = L 2 M 1 and M 2 exist by definition of L 1 and L 2 in LFSA Construct FSA M 3 from FSA’s M 1 and M 2 Argue L(M 3 ) = L 1 intersect L 2 There exists FSA M 3 s.t. L(M 3 ) = L 1 intersect L 2 L 1 intersect L 2 is in LFSA L 1 intersect L 2 L1L1 L2L2 LFSA M3M3 M1M1 M2M2 FSA’s

12 Algorithm Specification Input –Two FSA’s M 1 and M 2 Output –FSA M 3 such that L(M 3 ) = L(M 1 ) intersection L(M 2 ) FSA M 1 FSA M 2 FSA M 3 Alg

13 Use Old Ideas Key concept: Try ideas from previous closure property proofs Example –How did the algorithm for the intersection of two solvable languages is a solvable language work? Run both input programs on the input string Say yes only if both say yes –Try to create an FSA M 3 that simultaneously runs M 1 and M 2 on the input string FSA M 1 FSA M 2 FSA M 3 Alg

14 1 Run M 1 and M 2 Simultaneously M1M1 A B 0 10,1 M2M2,A 0,A1,A2,A,B 0,B1,B2,B M3M3 What happens when M 1 and M 2 run on input string 11010? FSA M 1 FSA M 2 FSA M 3 Alg

15 Construction Input –FSA M 1 = (Q 1,  , q 1,  , A 1 ) –FSA M 2 = (Q 2,  , q 2,  , A 2 ) Output –FSA M 3 = (Q 3,  , q 3,  , A 3 ) –What is Q 3 ? Q 3 = Q 1 X Q 2 where X is cartesian product In this case, Q 3 = {(,A), (,B), (0,A), (0,B), (1,A), (1,B), (2,A), (2,B)} –What is  3 ?  3 =  1 =  2 In this case,  3 = {0,1} M1M1 A B 0 10,1 M2M2

16 Construction Input –FSA M 1 = (Q 1,  , q 1,  , A 1 ) –FSA M 2 = (Q 2,  , q 2,  , A 2 ) Output –FSA M 3 = (Q 3,  , q 3,  , A 3 ) –What is q 3 ? q 3 = (q 1, q 2 ) In this case, q 3 = (,A) –What is A 3 ? A 3 = {(p, q) | p in A 1 and q in A 2 } In this case, A 3 = {(0,B)} M1M1 A B 0 10,1 M2M2

17 Construction Input –FSA M 1 = (Q 1,  , q 1,  , A 1 ) –FSA M 2 = (Q 2,  , q 2,  , A 2 ) Output –FSA M 3 = (Q 3,  , q 3,  , A 3 ) –What is  3 ? For all p in Q 1, q in Q 2, a in ,  3 ((p,q),a) = (  1 (p,a),  2 (q,a)) In this case, –  3 ((0,A),0) = (  1 (0,0),  2 (A,0)) – = (0,B) –  3 ((0,A),1) = (  1 (0,1),  2 (A,1)) – = (1,A) M1M1 A B 0 10,1 M2M2

18 Example Summary M1M1 A B 0 10,1 M2M2,A 0,A1,A2,A,B 0,B 1,B2,B M3M

19 Observation Input –FSA M 1 = (Q 1,  , q 1,  , A 1 ) –FSA M 2 = (Q 2,  , q 2,  , A 2 ) Output –FSA M 3 = (Q 3,  , q 3,  , A 3 ) –What is A 3 ? A 3 = {(p, q) | p in A 1 and q in A 2 } What if operation were different? –Set union, set difference, symmetric difference

20 Observation continued Input –FSA M 1 = (Q 1,  , q 1,  , A 1 ) –FSA M 2 = (Q 2,  , q 2,  , A 2 ) Output –FSA M 3 = (Q 3,  , q 3,  , A 3 ) –What is A 3 ? Set intersection: A 3 = {(p, q) | p in A 1 and q in A 2 } Set union: A 3 = {(p, q) | p in A 1 or q in A 2 } Set difference: A 3 = {(p, q) | p in A 1 and q not in A 2 } Symmetric difference: A 3 = {(p, q) | (p in A 1 and q not in A 2 ) or (p not in A 1 and q in A 2 ) }

21 Observation conclusion LFSA is closed under –set intersection –set union –set difference –symmetric difference The constructions used to prove these closure properties are essentially identical

22 Comments You should be able to execute this algorithm –Convert two FSA’s into a third FSA with the correct properties. You should understand the idea behind this algorithm –The third FSA essentially runs both input FSA’s simultaneously on any input string –How we set A 3 depending on the specific set operation You should understand the importance of this algorithm –Design tool Suppose you need to build an FSA to accept some language L 3 You observe L 3 = L 1 intersection L 2 You already have or can easily build FSA’s to accept L 1 and L 2 Use this algorithm on those FSA’s to constrct an FSA to accept L 3 You should be able to construct new algorithms for new closure property proofs

23 Comparison L 1 intersect L 2 L1L1 L2L2 LFSA M3M3 M1M1 M2M2 FSA’s LFSA REC FSA’s C++ Programs L L M P